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Let $K=\frac{\mathbb{Q}[x]}{<f(x)>}$ where $f(x)$ is irreducible over $\mathbb{Q}$ and has even degree. I want to find $K_2$ such that $ \mathbb{Q} \subseteq K_2\subseteq K$ and $[K_2:\mathbb{Q}]=2$.

If K is a cyclic Galois extension over $\mathbb{Q}$ then discriminant of $f(x)$ solves the problem.

But what if $K$ is not a cyclic galois extension?

Or even particular when $K$ is not even a Galois extension.

Improve this question
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    $\begingroup$ do you know or can calculate the galois group of (the splitting field of) f(x)? $\endgroup$
    – ricardorr
    Commented Aug 23, 2016 at 0:12
  • $\begingroup$ If degree of splitting field is small "poly(deg ($f$))" than we can calculate the galois group of the splitting field, but its degree two subfield might not be a subfield of $K$ as well. $\endgroup$
    – xyz
    Commented Aug 23, 2016 at 3:11
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    $\begingroup$ How does the discriminant of $f$ solve the problem when the extension is Galois? $\mathbb{Q}(\sqrt{\Delta_f})$ works when the discriminant isn't a square in $\mathbb{Q}$, but that needn't always be the case. In fact, if $K/\mathbb{Q}$ is a Galois extension with Galois group $A_n$ ($n\ge 4$), there are no index-2 subgroups, so $K_2$ won't even exist. Is it just the case that there is no $K_2$ when the discriminant is a square? $\endgroup$
    – Andrew Morgan
    Commented Aug 27, 2016 at 5:11
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    $\begingroup$ To answer my own question at the end of my previous comment: Consider $f(x) = x^4 + 1$. It's irreducible with roots $\pm\sqrt{\pm i}$. It has discriminant 256, which is a square, but its splitting field contains the degree-2 field $\mathbb{Q}(i)$. $\endgroup$
    – Andrew Morgan
    Commented Aug 27, 2016 at 5:28
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    $\begingroup$ For Galois extensions: It's known that the discriminant is a square iff the Galois group is a sub(-permutation-)group of $A_n$ (where $n$ is the degree of $f$). Hence if the Galois group isn't a subgroup of $A_n$, the discriminant method works. This generalizes the case of cyclic Galois extensions: Galois groups act transitively on the roots, so a cyclic Galois group has to be generated by an $n$-cycle; however, when $n$ is even, $n$-cycles are odd permutations, and hence aren't in $A_n$. $\endgroup$
    – Andrew Morgan
    Commented Aug 27, 2016 at 6:35

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